How do you solve the triangle given #B=75^circ20', a=6.2, c=9.5#?

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Answer 1 · 2017-05-27T14:12:49

#angleB=14°40',angleA=75°20',b=1.623#

angle B in the question should be angle A

#angleA=75°20',a=6.2,c=9.5#

#:.#Ajacent/opposite#=b/a= Cot 75°20'#

#:.b/6.2= Cot 75°20'#

multiply both sides by 6.2

#:.b=Cot 75°20' xx 6.2#

#:.b=0.261723366 xx 6.2#

#:.b=1.622684873#

#:.color(blue)(b=1.623# to the nearest 3 decimal places.

#:.180°-(90°+75°20')=14°40'#

#:.color(blue)(angleB=14°40'#